Question: Solve for $n$, $ \dfrac{2n + 2}{n + 3} = \dfrac{1}{6} $
Solution: Multiply both sides of the equation by $n + 3$ $ 2n + 2 = \dfrac{n + 3}{6} $ Multiply both sides of the equation by $6$ $ 6(2n + 2) = n + 3 $ $12n + 12 = n + 3$ $11n + 12 = 3$ $11n = -9$ $n = -\dfrac{9}{11}$